Question bank/JEE Main/Mathematics

JEE MainMathematicsToughVectors

If
a→,b→ and c→\overset{\rightarrow}{a},\overset{\rightarrow}{b}\ and\ \overset{\rightarrow}{c}are
unit vectors, then

∣a→−b→∣2+∣b→−c→∣2+∣c→−a→∣2|\overset{\rightarrow}{a} - \overset{\rightarrow}{b}|^{2} + |\overset{\rightarrow}{b} - \overset{\rightarrow}{c}|^{2} + |\overset{\rightarrow}{c} - \overset{\rightarrow}{a}|^{2}does
NOT exceed

rks
∣a→−b→∣2+∣b→−c→∣2+∣c→−a→∣2|\overset{\rightarrow}{a} - \overset{\rightarrow}{b}|^{2} + |\overset{\rightarrow}{b} - \overset{\rightarrow}{c}|^{2} + |\overset{\rightarrow}{c} - \overset{\rightarrow}{a}|^{2} eku

Show answer and solution
∵(a→+b→+c→)2=Σa→2+2Σa→⋅b→≥0⇒2Σa→b→≥−3\because(\overset{\rightarrow}{a} + \overset{\rightarrow}{b} + \overset{\rightarrow}{c})^{2} = \Sigma{\overset{\rightarrow}{a}}^{2} + 2\Sigma\overset{\rightarrow}{a} \cdot \overset{\rightarrow}{b} \geq 0 \Rightarrow 2\Sigma\overset{\rightarrow}{a}\overset{\rightarrow}{b} \geq - 3
⇒Σ∣a→−b→∣2=2Σa→2−2Σa→b→⇒Σ∣a→−b→∣2⇒Σ∣a→−b→∣2≥9\Rightarrow \Sigma|\overset{\rightarrow}{a} - \overset{\rightarrow}{b}|^{2} = 2\Sigma{\overset{\rightarrow}{a}}^{2} - 2\Sigma\overset{\rightarrow}{a}\overset{\rightarrow}{b} \Rightarrow \Sigma|\overset{\rightarrow}{a} - \overset{\rightarrow}{b}|^{2} \Rightarrow \Sigma|\overset{\rightarrow}{a} - \overset{\rightarrow}{b}|^{2} \geq 9

Source: Grass-G / M-21 Mathematics Paper 19 Dec Hindi Eng.docx

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